Chapter 14 Simple Linear Regression - Seattle...

14 - 1

Chapter 14 Simple Linear Regression Learning Objectives 1. Understand how regression analysis can be used to develop an equation that estimates

mathematically how two variables are related. 2. Understand the differences between the regression model, the regression equation, and the estimated

regression equation. 3. Know how to fit an estimated regression equation to a set of sample data based upon the least-

squares method. 4. Be able to determine how good a fit is provided by the estimated regression equation and compute

the sample correlation coefficient from the regression analysis output. 5. Understand the assumptions necessary for statistical inference and be able to test for a significant

relationship. 6. Learn how to use a residual plot to make a judgement as to the validity of the regression

assumptions, recognize outliers, and identify influential observations. 7. Know how to develop confidence interval estimates of y given a specific value of x in both the case

of a mean value of y and an individual value of y. 8. Be able to compute the sample correlation coefficient from the regression analysis output. 9. Know the definition of the following terms: independent and dependent variable simple linear regression regression model regression equation and estimated regression equation scatter diagram coefficient of determination standard error of the estimate confidence interval prediction interval residual plot standardized residual plot outlier influential observation leverage

Chapter 14

14 - 2

Solutions: 1 a.

b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the

relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion.

d. Summations needed to compute the slope and y-intercept are: 215 40 ( )( ) 26 ( ) 10i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 262.6

10( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 8 2 6 3 0 2= − = − =( . )( ) .

$ . .y x= −0 2 2 6

e. $ . . ( ) .y = − =0 2 2 6 4 10 6

0

2

46

810

1214

16

0 1 2 3 4 5 6

x

y

Simple Linear Regression

14 - 3

2. a.



d. Summations needed to compute the slope and y-intercept are:

219 116 ( )( ) 57.8 ( ) 30.8i i i i ix y x x y y x xΣ = Σ = Σ − − = − Σ − =

1 2

( )( ) 57.81.8766

30.8( )i i

i

x x y yb

x xΣ − − −

= = = −Σ −

b y b x0 1 23 2 18766 38 303311= − = − − =. ( . )( . ) .

$ . .y x= −30 33 188

e. $ . . ( ) .y = − =3033 188 6 19 05

0

5

10

15

20

25

30

35

0 2 4 6 8 10

x

y

Chapter 14

14 - 4

3. a.

b. Summations needed to compute the slope and y-intercept are: 226 17 ( )( ) 11.6 ( ) 22.8i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 11.60.5088

22.8( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 34 05088 2 07542= − = − =. ( . )(5. ) .

$ . .y x= +0 75 0 51

c. $ . . ( ) .y = + =0 75 0 51 4 2 79

0

1

2

3

4

5

6

7

0 2 4 6 8 10

x

y


14 - 5

100

105

110

115

120

125

130

135

61 62 63 64 65 66 67 68 69

x

y

4. a.

b. There appears to be a linear relationship between c. Many different straight lines can be drawn to provide a linear approximation of the


d. Summations needed to compute the slope and y-intercept are: 2325 585 ( )( ) 110 ( ) 20i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 1105.5

20( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 117 5 65 240 5= − = − = −(5. )( ) .

$ . .y x= − +2405 55

e. $ . . . . ( )y x= − + = − + =2405 55 2405 55 63 106 pounds

Chapter 14

14 - 6

5. a.



Summations needed to compute the slope and y-intercept are: 2420.6 5958.7 ( )( ) 142,040.3443 ( ) 9847.6486i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 142,040.344314.4238

9847.6486( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 8512429 14 4238 60 0857 1542= − = − = −. ( . )( . ) .

$ . .y x= − +15 42 14 42

d. A one million dollar increase in media expenditures will increase case sales by approximately 14.42

million. e. $ . . . . ( ) .y x= − + = − + =15 42 14 42 1542 14 42 70 99398

100

300

500

700

900

1100

1300

1500

1700

1900

2100

0 20 40 60 80 100 120 140

x

y


14 - 7

6. a.

b. There appears to be a linear relationship between x and y. c. Summations needed to compute the slope and y-intercept are: 2667.2 7.18 ( )( ) 9.0623 ( ) 128.7i i i i ix y x x y y x xΣ = Σ = Σ − − = − Σ − =

1 2

( )( ) 9.06230.0704

128.7( )i i

i

x x y yb

x xΣ − − −

= = = −Σ −

b y b x0 1 0 7978 0 0704 741333 6 02= − = − − =. ( . )( . ) .

$ . .y x= −6 02 0 07

d. A one percent increase in the percentage of flights arriving on time will decrease the number of

complaints per 100,000 passengers by 0.07. e $ . . . . (80) .y x= − = − =6 02 0 07 602 0 07 042

0

0.2

0.4

0.6

0.8

1

1.2

1.4

66 68 70 72 74 76 78 80 82 84x

y

Chapter 14

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7. a. b. Let x = DJIA and y = S&P. Summations needed to compute the slope and y-intercept are: 2104,850 14,233 ( )( ) 268,921 ( ) 1,806,384i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 268,9210.14887

1,806,384( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 1423.3 (.14887)(10,485) 137.629b y b x= − = − = −

ˆ 137.63 0.1489y x= − +

c. ˆ 137.63 0.1489(11,000) 1500.27y = − + = or approximately 1500

8. a. Summations needed to compute the slope and y-intercept are: 2121 1120.9 ( )( ) 544.0429 ( ) 177.4286i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 544.04293.0663

177.4286( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 1601286 30663 17 2857 107 13= − = − =. ( . )( . ) .

$ . .y x= +107 13 3 07

b. Increasing the number of times an ad is aired by one will increase the number of household exposures

by approximately 3.07 million.

1300

1350

1400

1450

1500

1550

9600 9800 10000 10200 10400 10600 10800 11000 11200

DJIA

S&P


14 - 9

c. $ . . . . ( ) .y x= + = + =107 13 3 07 10713 307 15 1532

9. a.

b. Summations needed to compute the slope and y-intercept are: 270 1080 ( )( ) 568 ( ) 142i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 5684

142( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 108 4 7 80= − = − =( )( )

$y x= +80 4

c. $ ( )y x= + = + =80 4 80 4 9 116

50

60

70

80

90

100

110

120

130

140

150

0 2 4 6 8 10 12 14x

y

Chapter 14

14 - 10

10. a. b. Let x = performance score and y = overall rating. Summations needed to compute the slope and y-

intercept are: 22752 1177 ( )( ) 1723.73 ( ) 11,867.73i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 1723.730.1452

11,867.73( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 78.4667 (.1452)(183.4667) 51.82b y b x= − = − =

ˆ 51.82 0.145y x= +

c. ˆ 51.82 0.145(225) 84.4y = + = or approximately 84

60

65

70

75

80

85

90

95

100 150 200 250

Performance Score

Ove

rall

Rat

ing


14 - 11

11. a.

b. There appears to be a linear relationship between the variables. c. The summations needed to compute the slope and the y-intercept are: 22973.3 3925.6 ( )( ) 453,345.042 ( ) 483,507.581i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 453,345.0420.9385

483,507.581( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 392 56 09385 297 33 11352= − = − =. ( . )( . ) .

$ . .y x= +11352 0 94

d. $ . . . . (500) .y x= + = + =11352 0 94 11352 094 5835

0.0

100.0

200.0

300.0

400.0

500.0

600.0

700.0

800.0

900.0

0.0 100.0 200.0 300.0 400.0 500.0 600.0 700.0 800.0

x

y

Chapter 14

14 - 12

12. a.

b. There appears to be a positive linear relationship between the number of employees and the revenue. c. Let x = number of employees and y = revenue. Summations needed to compute the slope and y-

intercept are: 24200 1669 ( )( ) 4,658,594,168 ( ) 14,718,343,803i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 4,658,594,1680.316516

14,718,343,803( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 14,048 (.316516)(40,299) 1293b y b x= − = − =

ˆ 1293 0.3165y x= +

d. ˆ 1293 .3165(75,000) 25,031y = + =

0

5000

10000

15000

20000

25000

30000

35000

40000

0 20000 40000 60000 80000 100000

Number of Employees

Rev

enue


14 - 13

13. a.

b. The summations needed to compute the slope and the y-intercept are: 2399 97.1 ( )( ) 1233.7 ( ) 7648i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 1233.70.16131

7648( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 1387143 016131 4 67675= − = − =. ( . )(57) .

$ . .y x= +4 68 016

c. $ . . . . (52. ) .y x= + = + =4 68 016 4 68 016 5 1308 or approximately $13,080.

The agent's request for an audit appears to be justified.

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0.0 20.0 40.0 60.0 80.0 100.0 120.0 140.0

x

y

Chapter 14

14 - 14

14. a.

b. The summations needed to compute the slope and the y-intercept are:

21677.25 1404.3 ( )( ) 897.9493 ( ) 3657.4568i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 897.94930.2455

3657.4568( )i i

i

x x y yb

x xΣ − −

= = =Σ −

b y b x0 1 70 215 0 2455 8625 49 63= − = − =. ( . )(83. ) .

$ . .y x= +49 63 2455

c. $ . . . . (80) .y x= + = + =49 63 2455 49 63 2455 69 3%

15. a. The estimated regression equation and the mean for the dependent variable are: $ . .y x yi i= + =02 2 6 8

The sum of squares due to error and the total sum of squares are SSE SST= ∑ − = = ∑ − =( $ ) . ( )y y y yi i i

2 212 40 80

Thus, SSR = SST - SSE = 80 - 12.4 = 67.6 b. r2 = SSR/SST = 67.6/80 = .845 The least squares line provided a very good fit; 84.5% of the variability in y has been explained by the

least squares line. c. r = = +. .845 9192

60

65

70

75

80

85

60 70 80 90 100 110x

y


14 - 15

16. a. The estimated regression equation and the mean for the dependent variable are:

ˆ 30.33 1.88 23.2iy x y= − =

The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 6.33 SST ( ) 114.80i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47 b. r2 = SSR/SST = 108.47/114.80 = .945 The least squares line provided an excellent fit; 94.5% of the variability in y has been explained by the

estimated regression equation. c. r = = −. .945 9721 Note: the sign for r is negative because the slope of the estimated regression equation is negative. (b1 = -1.88)

17. The estimated regression equation and the mean for the dependent variable are: ˆ .75 .51 3.4iy x y= + =

The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 5.3 SST ( ) 11.2i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9 r2 = SSR/SST = 5.9/11.2 = .527 We see that 52.7% of the variability in y has been explained by the least squares line. r = = +. .527 7259 18. a. The estimated regression equation and the mean for the dependent variable are: ˆ 1790.5 581.1 3650y x y= + =

The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 85,135.14 SST ( ) 335,000i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 335,000 - 85,135.14 = 249,864.86 b. r2 = SSR/SST = 249,864.86/335,000 = .746 We see that 74.6% of the variability in y has been explained by the least squares line.

Chapter 14

14 - 16

c. r = = +. .746 8637 19. a. The estimated regression equation and the mean for the dependent variable are: ˆ 137.63 .1489 1423.3y x y= − + =

The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 7547.14 SST ( ) 47,582.10i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 47,582.10 - 7547.14 = 40,034.96 b. r2 = SSR/SST = 40,034.96/47,582.10 = .84 We see that 84% of the variability in y has been explained by the least squares line. c. .84 .92r = = + 20. a. Let x = income and y = home price. Summations needed to compute the slope and y-intercept are: 21424 2455.5 ( )( ) 4011 ( ) 1719.618i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 40112.3325

1719.618( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 136.4167 (2.3325)(79.1111) 48.11b y b x= − = − = −

ˆ 48.11 2.3325y x= − +

b. The sum of squares due to error and the total sum of squares are 2 2ˆSSE ( ) 2017.37 SST ( ) 11,373.09i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 11,373.09 – 2017.37 = 9355.72 r2 = SSR/SST = 9355.72/11,373.09 = .82 We see that 82% of the variability in y has been explained by the least squares line. .82 .91r = = + c. ˆ 48.11 2.3325(95) 173.5y = − + = or approximately $173,500

21. a. The summations needed in this problem are: 23450 33,700 ( )( ) 712,500 ( ) 93,750i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 712,5007.6

93,750( )i i

i

x x y yb

x xΣ − −

= = =Σ −


14 - 17

0 1 5616.67 (7.6)(575) 1246.67b y b x= − = − =

$ . .y x= +1246 67 7 6

b. $7.60 c. The sum of squares due to error and the total sum of squares are: 2 2ˆSSE ( ) 233,333.33 SST ( ) 5,648,333.33i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 5,648,333.33 - 233,333.33 = 5,415,000 r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587 We see that 95.87% of the variability in y has been explained by the estimated regression equation. d. $ . . . . (500) $5046.y x= + = + =1246 67 7 6 1246 67 7 6 67

22. a. The summations needed in this problem are: 2613.1 70 ( )( ) 5766.7 ( ) 45,833.9286i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 5766.70.1258

45,833.9286( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 10 (0.1258)(87.5857) 1.0183b y b x= − = − = −

ˆ 1.0183 0.1258y x= − +

b. The sum of squares due to error and the total sum of squares are: 2 2ˆSSE ( ) 1272.4495 SST ( ) 1998i i iy y y y= ∑ − = = ∑ − =

Thus, SSR = SST - SSE = 1998 - 1272.4495 = 725.5505 r2 = SSR/SST = 725.5505/1998 = 0.3631 Approximately 37% of the variability in change in executive compensation is explained by the two-

year change in the return on equity. c. r = = +0 3631 0 6026. . It reflects a linear relationship that is between weak and strong. 23. a. s2 = MSE = SSE / (n - 2) = 12.4 / 3 = 4.133 b. s = = =MSE 4 133 2 033. . c. 2( ) 10ix xΣ − =

Chapter 14

14 - 18

1 2

2.0330.643

10( )b

i

ss

x x= = =

Σ −

d. t bsb

= = =1

1

2 6643

4 04..

.

t.025 = 3.182 (3 degrees of freedom) Since t = 4.04 > t.05 = 3.182 we reject H0: β1 = 0 e. MSR = SSR / 1 = 67.6 F = MSR / MSE = 67.6 / 4.133 = 16.36 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 16.36 > F.05 = 10.13 we reject H0: β1 = 0

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 67.6 1 67.6 16.36 Error 12.4 3 4.133 Total 80.0 4

24. a. s2 = MSE = SSE / (n - 2) = 6.33 / 3 = 2.11 b. s = = =MSE 2 11 1453. . c. 2( ) 30.8ix xΣ − =

1 2

1.4530.262

30.8( )b

i

ss

x x= = =

Σ −

d. t bsb

= = − = −1

1

188262

718..

.

t.025 = 3.182 (3 degrees of freedom) Since t = -7.18 < -t.025 = -3.182 we reject H0: β1 = 0 e. MSR = SSR / 1 = 8.47 F = MSR / MSE = 108.47 / 2.11 = 51.41 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 51.41 > F.05 = 10.13 we reject H0: β1 = 0

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 108.47 1 108.47 51.41 Error 6.33 3 2.11 Total 114.80 4


14 - 19

25. a. s2 = MSE = SSE / (n - 2) = 5.30 / 3 = 1.77 s = = =MSE 177 133. . b. 2( ) 22.8ix xΣ − =

1 2

1.330.28

22.8( )b

i

ss

x x= = =

Σ −

t bsb

= = =1

1

5128

182..

.

t.025 = 3.182 (3 degrees of freedom) Since t = 1.82 < t.025 = 3.182 we cannot reject H0: β1 = 0; x and y do not appear to be related. c. MSR = SSR/1 = 5.90 /1 = 5.90 F = MSR/MSE = 5.90/1.77 = 3.33 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 3.33 < F.05 = 10.13 we cannot reject H0: β1 = 0; x and y do not appear to be related. 26. a. s2 = MSE = SSE / (n - 2) = 85,135.14 / 4 = 21,283.79 s = = =MSE 21 283 79 14589, . .

2( ) 0.74ix xΣ − =

1 2

145.89169.59

0.74( )b

i

ss

x x= = =

Σ −

t bsb

= = =1

1

58108169 59

3 43..

.

t.025 = 2.776 (4 degrees of freedom) Since t = 3.43 > t.025 = 2.776 we reject H0: β1 = 0 b. MSR = SSR / 1 = 249,864.86 / 1 = 249.864.86 F = MSR / MSE = 249,864.86 / 21,283.79 = 11.74 F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 11.74 > F.05 = 7.71 we reject H0: β1 = 0 c.

Chapter 14

14 - 20

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 249864.86 1 249864.86 11.74 Error 85135.14 4 21283.79 Total 335000 5

27. The sum of squares due to error and the total sum of squares are: SSE = 2ˆ( ) 170i iy yΣ − = SST = 2442

Thus, SSR = SST - SSE = 2442 - 170 = 2272 MSR = SSR / 1 = 2272 SSE = SST - SSR = 2442 - 2272 = 170 MSE = SSE / (n - 2) = 170 / 8 = 21.25 F = MSR / MSE = 2272 / 21.25 = 106.92 F.05 = 5.32 (1 degree of freedom numerator and 8 denominator) Since F = 106.92 > F.05 = 5.32 we reject H0: β1 = 0. Years of experience and sales are related. 28. SST = 411.73 SSE = 161.37 SSR = 250.36 MSR = SSR / 1 = 250.36 MSE = SSE / (n - 2) = 161.37 / 13 = 12.413 F = MSR / MSE = 250.36 / 12.413= 20.17 F.05 = 4.67 (1 degree of freedom numerator and 13 denominator) Since F = 20.17 > F.05 = 4.67 we reject H0: β1 = 0. 29. SSE = 233,333.33 SST = 5,648,333.33 SSR = 5,415,000 MSE = SSE/(n - 2) = 233,333.33/(6 - 2) = 58,333.33 MSR = SSR/1 = 5,415,000 F = MSR / MSE = 5,415,000 / 58,333.25 = 92.83

Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 5,415,000.00 1 5,415,000 92.83 Error 233,333.33 4 58,333.33 Total 5,648,333.33 5

F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 92.83 > 7.71 we reject H0: β1 = 0. Production volume and total cost are related.


14 - 21

30. Using the computations from Exercise 22, SSE = 1272.4495 SST = 1998 SSR = 725.5505 s = =254 4899 1595. . 2( )ix x∑ − = 45,833.9286

1 2

15.950.0745

45,833.9286( )b

i

ss

x x= = =

Σ −

tbsb

= = =1

1

012580 0745

169..

.

t.025 = 2.571 Since t = 1.69 < 2.571, we cannot reject H0: β1 = 0 There is no evidence of a significant relationship between x and y. 31. SST = 11,373.09 SSE = 2017.37 SSR = 9355.72 MSR = SSR / 1 = 9355.72 MSE = SSE / (n - 2) = 2017.37/ 16 = 126.0856 F = MSR / MSE = 9355.72/ 126.0856 = 74.20 F.01 = 8.53 (1 degree of freedom numerator and 16 denominator) Since F = 74.20 > F.01 = 8.53 we reject H0: β1 = 0. 32. a. s = 2.033 23 ( ) 10ix x x= Σ − =

p

2 2p

ˆ 2

( )1 1 (4 3)2.033 1.115 10( )y

i

x xs s

n x x

− −= + = + =Σ −

b. $ . . . . ( ) .y x= + = + =0 2 2 6 0 2 2 6 4 106

$

/ $y t syp p± α 2

10.6 ± 3.182 (1.11) = 10.6 ± 3.53 or 7.07 to 14.13

c. 2 2

pind 2

( )1 1 (4 3)1 2.033 1 2.325 10( )i

x xs s

n x x

− −= + + = + + =Σ −

Chapter 14

14 - 22

d. $ /y t sp ind± α 2

10.6 ± 3.182 (2.32) = 10.6 ± 7.38 or 3.22 to 17.98 33. a. s = 1.453 b. 23.8 ( ) 30.8ix x x= Σ − =

p

2 2p

ˆ 2

( )1 1 (3 3.8)1.453 .0685 30.8( )y

i

x xs s

n x x

− −= + = + =Σ −

$ . . . . ( ) .y x= − = − =3033 188 30 33 188 3 24 69

$

/ $y t syp p± α 2

24.69 ± 3.182 (.68) = 24.69 ± 2.16 or 22.53 to 26.85

c. 2 2

pind 2

( )1 1 (3 3.8)1 1.453 1 1.615 30.8( )i

x xs s

n x x

− −= + + = + + =Σ −

d. $ /y t sp ind± α 2

24.69 ± 3.182 (1.61) = 24.69 ± 5.12 or 19.57 to 29.81 34. s = 1.33 25.2 ( ) 22.8ix x x= Σ − =

p

2 2p

ˆ 2

( )1 1 (3 5.2)1.33 0.855 22.8( )y

i

x xs s

n x x

− −= + = + =Σ −

$ . . . . ( ) .y x= + = + =0 75 0 51 0 75 051 3 2 28

$

/ $y t syp p± α 2

2.28 ± 3.182 (.85) = 2.28 ± 2.70 or -.40 to 4.98


14 - 23

2 2

pind 2

( )1 1 (3 5.2)1 1.33 1 1.585 22.8( )i

x xs s

n x x

− −= + + = + + =Σ −

$ /y t sp ind± α 2

2.28 ± 3.182 (1.58) = 2.28 ± 5.03 or -2.27 to 7.31 35. a. s = 145.89 23.2 ( ) 0.74ix x x= Σ − =

p

2 2p

ˆ 2

( )1 1 (3 3.2)145.89 68.546 0.74( )y

i

x xs s

n x x

− −= + = + =Σ −

$ . . . . ( ) .y x= + = + =290 54 58108 29054 58108 3 203378

$

/ $y t syp p± α 2

2,033.78 ± 2.776 (68.54) = 2,033.78 ± 190.27 or $1,843.51 to $2,224.05

b. 2 2

pind 2

( )1 1 (3 3.2)1 145.89 1 161.196 0.74( )i

x xs s

n x x

− −= + + = + + =Σ −


2,033.78 ± 2.776 (161.19) = 2,033.78 ± 447.46 or $1,586.32 to $2,481.24 36. a. ˆ 51.819 .1452 51.819 .1452(200) 80.859y x= + = + =

b. s = 3.5232 2183.4667 ( ) 11,867.73ix x x= Σ − =

p

2 2p

ˆ 2

( )1 1 (200 183.4667)3.5232 1.05515 11,867.73( )y

i

x xs s

n x x

− −= + = + =Σ −

$

/ $y t syp p± α 2

80.859 ± 2.160 (1.055) = 80.859 ± 2.279

Chapter 14

14 - 24

or 78.58 to 83.14

c. 2 2

pind 2

( )1 1 (200 183.4667)1 3.5232 1 3.67815 11,867.73( )i

x xs s

n x x

− −= + + = + + =Σ −


80.859 ± 2.160 (3.678) = 80.859 ± 7.944 or 72.92 to 88.80 37. a. 257 ( ) 7648ix x x= Σ − =

s2 = 1.88 s = 1.37

p

2 2p

ˆ 2

( )1 1 (52.5 57)1.37 0.527 7648( )y

i

x xs s

n x x

− −= + = + =Σ −

$

/ $y t syp p± α 2

13.08 ± 2.571 (.52) = 13.08 ± 1.34 or 11.74 to 14.42 or $11,740 to $14,420 b. sind = 1.47 13.08 ± 2.571 (1.47) = 13.08 ± 3.78 or 9.30 to 16.86 or $9,300 to $16,860 c. Yes, $20,400 is much larger than anticipated. d. Any deductions exceeding the $16,860 upper limit could suggest an audit. 38. a. $ . . (500) $5046.y = + =1246 67 7 6 67

b. 2575 ( ) 93,750ix x x= Σ − =

s2 = MSE = 58,333.33 s = 241.52

2 2

pind 2

( )1 1 (500 575)1 241.52 1 267.506 93,750( )i

x xs s

n x x

− −= + + = + + =Σ −


5046.67 ± 4.604 (267.50) = 5046.67 ± 1231.57


14 - 25

or $3815.10 to $6278.24 c. Based on one month, $6000 is not out of line since $3815.10 to $6278.24 is the prediction interval.

However, a sequence of five to seven months with consistently high costs should cause concern. 39. a. Summations needed to compute the slope and y-intercept are: 2227 2281.7 ( )( ) 6003.41 ( ) 1032.1i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 6003.415.816694

1032.1( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 228.17 (5.816694)(27.7) 67.047576b y b x= − = − =

$ . .y x= +67 0476 58167

b. SST = 39,065.14 SSE = 4145.141 SSR = 34,920.000 r2 = SSR/SST = 34,920.000/39,065.141 = 0.894 The estimated regression equation explained 89.4% of the variability in y; a very good fit. c. s2 = MSE = 4145.141/8 = 518.143 518.143 22.76s = =

p

2 2p

ˆ 2

( )1 1 (35 27.7)22.76 8.8610 1032.1( )y

i

x xs s

n x x

− −= + = + =Σ −

$ . . . . ( ) .y x= + = + =67 0476 58167 67 0476 58167 35 270 63

$

/ $y t syp p± α 2

270.63 ± 2.262 (8.86) = 270.63 ± 20.04 or 250.59 to 290.67

d. 2 2

pind 2

( )1 1 (35 27.7)1 22.76 1 24.4210 1032.1( )i

x xs s

n x x

− −= + + = + + =Σ −


270.63 ± 2.262 (24.42) = 270.63 ± 55.24 or 215.39 to 325.87 40. a. 9 b. y = 20.0 + 7.21x

Chapter 14

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c. 1.3626 d. SSE = SST - SSR = 51,984.1 - 41,587.3 = 10,396.8 MSE = 10,396.8/7 = 1,485.3 F = MSR / MSE = 41,587.3 /1,485.3 = 28.00 F.05 = 5.59 (1 degree of freedom numerator and 7 denominator) Since F = 28 > F.05 = 5.59 we reject H0: B1 = 0. e. y = 20.0 + 7.21(50) = 380.5 or $380,500

41. a. y = 6.1092 + .8951x

b. 1

1 1 .8951 06.01

.149b

b Bt

s− −

= = =

t.025 = 2.306 (1 degree of freedom numerator and 8 denominator) Since t = 6.01 > t.025 = 2.306 we reject H0: B1 = 0 c. y = 6.1092 + .8951(25) = 28.49 or $28.49 per month

42 a. y = 80.0 + 50.0x

b. 30 c. F = MSR / MSE = 6828.6/82.1 = 83.17 F.05 = 4.20 (1 degree of freedom numerator and 28 denominator) Since F = 83.17 > F.05 = 4.20 we reject H0: B1 = 0. Branch office sales are related to the salespersons. d. y = 80 + 50 (12) = 680 or $680,000

43. a. The Minitab output is shown below:

The regression equation is Price = - 11.8 + 2.18 Income Predictor Coef SE Coef T P Constant -11.80 12.84 -0.92 0.380 Income 2.1843 0.2780 7.86 0.000 S = 6.634 R-Sq = 86.1% R-Sq(adj) = 84.7%


14 - 27

Analysis of Variance Source DF SS MS F P Regression 1 2717.9 2717.9 61.75 0.000 Residual Error 10 440.1 44.0 Total 11 3158.0 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 75.79 2.47 ( 70.29, 81.28) ( 60.02, 91.56)

b. r2 = .861. The least squares line provided a very good fit. c. The 95% confidence interval is 70.29 to 81.28 or $70,290 to $81,280. d. The 95% prediction interval is 60.02 to 91.56 or $60,020 to $91,560. 44. a/b. The scatter diagram shows a linear relationship between the two variables. c. The Minitab output is shown below:

The regression equation is Rental$ = 37.1 - 0.779 Vacancy% Predictor Coef SE Coef T P Constant 37.066 3.530 10.50 0.000 Vacancy% -0.7791 0.2226 -3.50 0.003 S = 4.889 R-Sq = 43.4% R-Sq(adj) = 39.8% Analysis of Variance Source DF SS MS F P Regression 1 292.89 292.89 12.26 0.003 Residual Error 16 382.37 23.90 Total 17 675.26 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 17.59 2.51 ( 12.27, 22.90) ( 5.94, 29.23) 2 28.26 1.42 ( 25.26, 31.26) ( 17.47, 39.05) Values of Predictors for New Observations New Obs Vacancy% 1 25.0 2 11.3

d. Since the p-value = 0.003 is less than α = .05, the relationship is significant. e. r2 = .434. The least squares line does not provide a very good fit.

Chapter 14

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f. The 95% confidence interval is 12.27 to 22.90 or $12.27 to $22.90. g. The 95% prediction interval is 17.47 to 39.05 or $17.47 to $39.05. 45. a. 214 76 ( )( ) 200 ( ) 126i i i i ix y x x y y x xΣ = Σ = Σ − − = Σ − =

1 2

( )( ) 2001.5873

126( )i i

i

x x y yb

x xΣ − −

= = =Σ −

0 1 15.2 (1.5873)(14) 7.0222b y b x= − = − = −

ˆ 7.02 1.59y x= − +

b. The residuals are 3.48, -2.47, -4.83, -1.6, and 5.22 c.

-6

-4

-2

0

2

4

6

0 5 10 15 20 25

x

Res

idua

ls

With only 5 observations it is difficult to determine if the assumptions are satisfied.

However, the plot does suggest curvature in the residuals that would indicate that the error term assumptions are not satisfied. The scatter diagram for these data also indicates that the underlying relationship between x and y may be curvilinear.

d. 2 23.78s =

2 2

2

( ) ( 14)1 15 126( )

i ii

i

x x xh

n x x− −

= + = +Σ −

The standardized residuals are 1.32, -.59, -1.11, -.40, 1.49.


14 - 29

e. The standardized residual plot has the same shape as the original residual plot. The curvature observed indicates that the assumptions regarding the error term may not be satisfied.

46. a. ˆ 2.32 .64y x= +

b.

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-3-2-101

234

0 2 4 6 8 10

x

Res

idua

ls

The assumption that the variance is the same for all values of x is questionable. The variance appears

to increase for larger values of x. 47. a. Let x = advertising expenditures and y = revenue ˆ 29.4 1.55y x= +

b. SST = 1002 SSE = 310.28 SSR = 691.72 MSR = SSR / 1 = 691.72 MSE = SSE / (n - 2) = 310.28/ 5 = 62.0554 F = MSR / MSE = 691.72/ 62.0554= 11.15 F.05 = 6.61 (1 degree of freedom numerator and 5 denominator) Since F = 11.15 > F.05 = 6.61 we conclude that the two variables are related. c.

Chapter 14

14 - 30

-15

-10

-5

0

5

10

25 35 45 55 65

Predicted Values

Res

idua

ls

d. The residual plot leads us to question the assumption of a linear relationship between x and y. Even

though the relationship is significant at the .05 level of significance, it would be extremely dangerous to extrapolate beyond the range of the data.

48. a. ˆ 80 4y x= +


14 - 31

-8

-6

-4

-2

0

2

4

6

8

0 2 4 6 8 10 12 14

x

Res

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ls

b. The assumptions concerning the error term appear reasonable. 49. a. Let x = return on investment (ROE) and y = price/earnings (P/E) ratio. ˆ 32.13 3.22y x= − +

b.

-1.5

-1

-0.5

0

0.5

1

1.5

2

0 10 20 30 40 50 60

x

Stan

dard

ized

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ls

c. There is an unusual trend in the residuals. The assumptions concerning the error term appear

questionable.

Chapter 14

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50. a. The MINITAB output is shown below:

The regression equation is Y = 66.1 + 0.402 X Predictor Coef Stdev t-ratio p Constant 66.10 32.06 2.06 0.094 X 0.4023 0.2276 1.77 0.137 s = 12.62 R-sq = 38.5% R-sq(adj) = 26.1% Analysis of Variance SOURCE DF SS MS F p Regression 1 497.2 497.2 3.12 0.137 Error 5 795.7 159.1 Total 6 1292.9 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 1 135 145.00 120.42 4.87 24.58 2.11R R denotes an obs. with a large st. resid.

The standardized residuals are: 2.11, -1.08, .14, -.38, -.78, -.04, -.41 The first observation appears to be an outlier since it has a large standardized residual. b.

2.4+ - * STDRESID- - - 1.2+ - - - - * 0.0+ * - - * * - * - -1.2+ * - --+---------+---------+---------+---------+---------+----YHAT 110.0 115.0 120.0 125.0 130.0 135.0

The standardized residual plot indicates that the observation x = 135,y = 145 may be an outlier; note that this observation has a standardized residual of 2.11.


14 - 33

c. The scatter diagram is shown below

- Y - * - - 135+ - - * * - - 120+ * * - - - * - 105+ - - * ----+---------+---------+---------+---------+---------+--X 105 120 135 150 165 180

The scatter diagram also indicates that the observation x = 135,y = 145 may be an outlier; the implication is that for simple linear regression an outlier can be identified by looking at the scatter diagram.


The regression equation is Y = 13.0 + 0.425 X Predictor Coef Stdev t-ratio p Constant 13.002 2.396 5.43 0.002 X 0.4248 0.2116 2.01 0.091 s = 3.181 R-sq = 40.2% R-sq(adj) = 30.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 40.78 40.78 4.03 0.091 Error 6 60.72 10.12 Total 7 101.50 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 7 12.0 24.00 18.10 1.20 5.90 2.00R 8 22.0 19.00 22.35 2.78 -3.35 -2.16RX R denotes an obs. with a large st. resid. X denotes an obs. whose X value gives it large influence.

Chapter 14

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The standardized residuals are: -1.00, -.41, .01, -.48, .25, .65, -2.00, -2.16 The last two observations in the data set appear to be outliers since the standardized residuals for

these observations are 2.00 and -2.16, respectively.

b. Using MINITAB, we obtained the following leverage values: .28, .24, .16, .14, .13, .14, .14, .76 MINITAB identifies an observation as having high leverage if hi > 6/n; for these data, 6/n = 6/8

= .75. Since the leverage for the observation x = 22, y = 19 is .76, MINITAB would identify observation 8 as a high leverage point. Thus, we conclude that observation 8 is an influential observation.

c.

24.0+ * - Y - - - 20.0+ * - * - * - - 16.0+ * - * - - * - 12.0+ * - +---------+---------+---------+---------+---------+------X 0.0 5.0 10.0 15.0 20.0 25.0

The scatter diagram indicates that the observation x = 22, y = 19 is an influential observation. 52. a. The Minitab output is shown below:

The regression equation is Amount = 4.09 + 0.196 MediaExp Predictor Coef SE Coef T P Constant 4.089 2.168 1.89 0.096 MediaExp 0.19552 0.03635 5.38 0.001 S = 5.044 R-Sq = 78.3% R-Sq(adj) = 75.6% Analysis of Variance Source DF SS MS F P Regression 1 735.84 735.84 28.93 0.001


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Residual Error 8 203.51 25.44 Total 9 939.35 Unusual Observations Obs MediaExp Amount Fit SE Fit Residual St Resid 1 120 36.30 27.55 3.30 8.75 2.30R R denotes an observation with a large standardized residual

b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider

observation 1 to be an outlier. 53. a. The Minitab output is shown below:

The regression equation is Exposure = - 8.6 + 7.71 Aired Predictor Coef SE Coef T P Constant -8.55 21.65 -0.39 0.703 Aired 7.7149 0.5119 15.07 0.000 S = 34.88 R-Sq = 96.6% R-Sq(adj) = 96.2% Analysis of Variance Source DF SS MS F P Regression 1 276434 276434 227.17 0.000 Residual Error 8 9735 1217 Total 9 286169 Unusual Observations Obs Aired Exposure Fit SE Fit Residual St Resid 1 95.0 758.8 724.4 32.0 34.4 2.46RX R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence.

b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider

observation 1 to be an outlier. Minitab also identifies observation 1 as an influential observation. 54. a. The Minitab output is shown below:

The regression equation is Salary = 707 + 0.00482 MktCap Predictor Coef SE Coef T P Constant 707.0 118.0 5.99 0.000 MktCap 0.0048154 0.0008076 5.96 0.000 S = 379.8 R-Sq = 66.4% R-Sq(adj) = 64.5% Analysis of Variance

Chapter 14

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Source DF SS MS F P Regression 1 5129071 5129071 35.55 0.000 Residual Error 18 2596647 144258 Total 19 7725718 Unusual Observations Obs MktCap Salary Fit SE Fit Residual St Resid 6 507217 3325.0 3149.5 338.6 175.5 1.02 X 17 120967 116.2 1289.5 86.4 -1173.3 -3.17R R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence.

b. Minitab identifies observation 6 as having a large standardized residual and observation 17 as an

observation whose x value gives it large influence. A standardized residual plot against the predicted values is shown below:

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-1

0

1

2

3

500 1000 1500 2000 2500 3000 3500

Predicted Values

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55. No. Regression or correlation analysis can never prove that two variables are casually related. 56. The estimate of a mean value is an estimate of the average of all y values associated with the same x.

The estimate of an individual y value is an estimate of only one of the y values associated with a particular x.

57. To determine whether or not there is a significant relationship between x and y. However, if we reject

B1 = 0, it does not imply a good fit. 58. a. The Minitab output is shown below:

The regression equation is Price = 9.26 + 0.711 Shares Predictor Coef SE Coef T P Constant 9.265 1.099 8.43 0.000 Shares 0.7105 0.1474 4.82 0.001 S = 1.419 R-Sq = 74.4% R-Sq(adj) = 71.2% Analysis of Variance


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Source DF SS MS F P Regression 1 46.784 46.784 23.22 0.001 Residual Error 8 16.116 2.015 Total 9 62.900

b. Since the p-value corresponding to F = 23.22 = .001 < α = .05, the relationship is significant. c. 2r = .744; a good fit. The least squares line explained 74.4% of the variability in Price. d. ˆ 9.26 .711(6) 13.53y = + =


The regression equation is Options = - 3.83 + 0.296 Common Predictor Coef SE Coef T P Constant -3.834 5.903 -0.65 0.529 Common 0.29567 0.02648 11.17 0.000 S = 11.04 R-Sq = 91.9% R-Sq(adj) = 91.2% Analysis of Variance Source DF SS MS F P Regression 1 15208 15208 124.72 0.000 Residual Error 11 1341 122 Total 12 16550

b. ˆ 3.83 .296(150) 40.57y = − + = ; approximately 40.6 million shares of options grants outstanding.

c. 2r = .919; a very good fit. The least squares line explained 91.9% of the variability in Options. 60. a. The Minitab output is shown below:

The regression equation is IBM = 0.275 + 0.950 S&P 500 Predictor Coef StDev T P Constant 0.2747 0.9004 0.31 0.768 S&P 500 0.9498 0.3569 2.66 0.029 S = 2.664 R-Sq = 47.0% R-Sq(adj) = 40.3% Analysis of Variance Source DF SS MS F P Regression 1 50.255 50.255 7.08 0.029 Error 8 56.781 7.098

Chapter 14

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Total 9 107.036 b. Since the p-value = 0.029 is less than α = .05, the relationship is significant. c. r2 = .470. The least squares line does not provide a very good fit. d. Woolworth has higher risk with a market beta of 1.25. 61. a.

40

50

60

70

80

90

100

35 45 55 65 75 85

Low Temperature

Hig

h T

empe

ratu

re

b. It appears that there is a positive linear relationship between the two variables. c. The Minitab output is shown below:

The regression equation is High = 23.9 + 0.898 Low Predictor Coef SE Coef T P Constant 23.899 6.481 3.69 0.002 Low 0.8980 0.1121 8.01 0.000 S = 5.285 R-Sq = 78.1% R-Sq(adj) = 76.9% Analysis of Variance Source DF SS MS F P Regression 1 1792.3 1792.3 64.18 0.000


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Residual Error 18 502.7 27.9 Total 19 2294.9

d. Since the p-value corresponding to F = 64.18 = .000 < α = .05, the relationship is significant. e. 2r = .781; a good fit. The least squares line explained 78.1% of the variability in high temperature. f. .781 .88r = = +

62. The MINITAB output is shown below:

The regression equation is Y = 10.5 + 0.953 X Predictor Coef Stdev t-ratio p Constant 10.528 3.745 2.81 0.023 X 0.9534 0.1382 6.90 0.000 s = 4.250 R-sq = 85.6% R-sq(adj) = 83.8% Analysis of Variance SOURCE DF SS MS F p Regression 1 860.05 860.05 47.62 0.000 Error 8 144.47 18.06 Total 9 1004.53 Fit Stdev.Fit 95% C.I. 95% P.I. 39.13 1.49 ( 35.69, 42.57) ( 28.74, 49.52)

a. y = 10.5 + .953 x

b. Since the p-value corresponding to F = 47.62 = .000 < α = .05, we reject H0: β1 = 0. c. The 95% prediction interval is 28.74 to 49.52 or $2874 to $4952 d. Yes, since the expected expense is $3913. 63. a. The Minitab output is shown below:

The regression equation is Defects = 22.2 - 0.148 Speed Predictor Coef SE Coef T P Constant 22.174 1.653 13.42 0.000 Speed -0.14783 0.04391 -3.37 0.028 S = 1.489 R-Sq = 73.9% R-Sq(adj) = 67.4% Analysis of Variance Source DF SS MS F P Regression 1 25.130 25.130 11.33 0.028 Residual Error 4 8.870 2.217

Chapter 14

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Total 5 34.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 14.783 0.896 ( 12.294, 17.271) ( 9.957, 19.608)

b. Since the p-value corresponding to F = 11.33 = .028 < α = .05, the relationship is significant. c. 2r = .739; a good fit. The least squares line explained 73.9% of the variability in the number of

defects. d. Using the Minitab output in part (a), the 95% confidence interval is 12.294 to 17.271. 64. a. There appears to be a negative linear relationship between distance to work and number of days

absent.

b. The MINITAB output is shown below:

The regression equation is Y = 8.10 - 0.344 X Predictor Coef Stdev t-ratio p Constant 8.0978 0.8088 10.01 0.000 X -0.34420 0.07761 -4.43 0.002 s = 1.289 R-sq = 71.1% R-sq(adj) = 67.5% Analysis of Variance SOURCE DF SS MS F p Regression 1 32.699 32.699 19.67 0.002 Error 8 13.301 1.663 Total 9 46.000 Fit Stdev.Fit 95% C.I. 95% P.I. 6.377 0.512 ( 5.195, 7.559) ( 3.176, 9.577)

c. Since the p-value corresponding to F = 419.67 is .002 < α = .05. We reject H0 : β1 = 0. d. r2 = .711. The estimated regression equation explained 71.1% of the variability in y; this is a

reasonably good fit. e. The 95% confidence interval is 5.195 to 7.559 or approximately 5.2 to 7.6 days. 65. a. Let X = the age of a bus and Y = the annual maintenance cost.

The MINITAB output is shown below:


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The regression equation is Y = 220 + 132 X Predictor Coef Stdev t-ratio p Constant 220.00 58.48 3.76 0.006 X 131.67 17.80 7.40 0.000 s = 75.50 R-sq = 87.3% R-sq(adj) = 85.7% Analysis of Variance SOURCE DF SS MS F p Regression 1 312050 312050 54.75 0.000 Error 8 45600 5700 Total 9 357650 Fit Stdev.Fit 95% C.I. 95% P.I. 746.7 29.8 ( 678.0, 815.4) ( 559.5, 933.9)

b. Since the p-value corresponding to F = 54.75 is .000 < α = .05, we reject H0: β1 = 0. c. r2 = .873. The least squares line provided a very good fit. d. The 95% prediction interval is 559.5 to 933.9 or $559.50 to $933.90 66. a. Let X = hours spent studying and Y = total points earned

The MINITAB output is shown below:

The regression equation is Y = 5.85 + 0.830 X Predictor Coef Stdev t-ratio p Constant 5.847 7.972 0.73 0.484 X 0.8295 0.1095 7.58 0.000 s = 7.523 R-sq = 87.8% R-sq(adj) = 86.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 3249.7 3249.7 57.42 0.000 Error 8 452.8 56.6 Total 9 3702.5 Fit Stdev.Fit 95% C.I. 95% P.I. 84.65 3.67 ( 76.19, 93.11) ( 65.35, 103.96)

b. Since the p-value corresponding to F = 57.42 is .000 < α = .05, we reject H0: β1 = 0. c. 84.65 points d. The 95% prediction interval is 65.35 to 103.96

Chapter 14

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The regression equation is Audit% = - 0.471 +0.000039 Income Predictor Coef SE Coef T P Constant -0.4710 0.5842 -0.81 0.431 Income 0.00003868 0.00001731 2.23 0.038 S = 0.2088 R-Sq = 21.7% R-Sq(adj) = 17.4% Analysis of Variance Source DF SS MS F P Regression 1 0.21749 0.21749 4.99 0.038 Residual Error 18 0.78451 0.04358 Total 19 1.00200 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 0.8828 0.0523 ( 0.7729, 0.9927) ( 0.4306, 1.3349)

b. Since the p-value = 0.038 is less than α = .05, the relationship is significant. c. r2 = .217. The least squares line does not provide a very good fit. d. The 95% confidence interval is .7729 to .9927.

Chapter 14 Simple Linear Regression - Seattle...

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